Chemistry help?

[rice4thewin]

Any smart chemist out there who want to tell me how to solve the last two problems. Also help on how i got the first one right would be helpful too, hahahaha I guessed.

[kennethnakasone]

Lol, nice guess. Anyway, if you still need the help:

 

At the beginning, you have two tubes. One has Cadmium and Nitrate dissolved in it. The other has Sodium and Sulfur dissolved in it. When you mixed them together, the yellow stuff formed. I forgot how you actually determine it, but let's be honest: it's yellow, so it's not that surprising that it's partially Cadmium (lol).

 

As for which ions remain in solution, it's just asking "what stuff is in the clear part (around the CdS)". Since you used up all Cadmium from the first tube and Sulfur from the second tube to make the yellow stuff, the clear area must be made of the stuff that's left over from either tube.

 

As for the last question, it's telling you to write an equation that only lists the stuff that "changed" as a result of the experiment. So, while the Nitrate and Sodium didn't change (they both stayed dissolved 'n whatnot), the Cadmium and Sulfur changed from "dissolved" into "solid". So, the net ionic equation would state be something like "Cadmium (dissolved) and Sulfur (dissolved) turned into Cadmium Sulfate (solid)" .

 

...of course, that's unless I'm remembering this completely wrong lol.

[Bsmith]

could, yes.

would? no, because this is obvious schoolwork.

[corrado33]

Easy stuff. Please excuse my lack of subscripts.

 

Cd(NO3)2 + Na2S -> CdS + 2(NaNO3)

 

CdS is the precipitate, the Ions left in solution assuming the reaction goes to completion and stuff is added in stoichiometric amounts are Na+ and (NO3)-

 

The net ionic equation is just the above equation with all of the "spectator" ions removed (so in this case the Na and NO3 because they don't "do" anything except hang around in solution.)

 

Therefore it'd be (Cd)2+ + (S)2- -> CdS